Lacks's Level 5

Author:
Lacks

Language:
C/C++

Upload:
2026-07-10 17:23

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Platform:
Windows

Difficulty:
3.3

Quality:
4.7

Arch:
x86-64

Downloads:
219

Size:
18.00 KB

Writeups:
1

Comments:
9

Description

This is level 5 im uploading 1 - 10 crack mes

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Comments (9)
Writeups (1)

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r8 on 2026-07-11 14:55: [Click to reveal]

dandelion_msla on 2026-07-11 15:21: 6052 / 1213 Victory!

Lover1500 on 2026-07-12 18:37: [Click to reveal]

Toji on 2026-07-13 10:51: [Click to reveal]

Abh0rrent on 2026-07-14 16:08: This one was very similar to the last and was very similar in difficulty. The math involving how key1 should validate itself and validate key2 was interesting especially while needing a specific value for key2 to properly decrypt the flag. "Victory!". Interestingly, I looked at all 2-6 digit numbers, and there are only about 230 and will properly validate for this! I don't really want to bother looking for larger numbers but I know there are.

Phantaso on 2026-07-21 15:03: if ( v16 != 0 && v16 == 17 * (v16 / 0x11) && v3 == (v16 ^ 0x1337) + 42 ) Part 1: 17 | Part 2: 4944

kotsuge on 2026-07-21 21:22: Part1 % 17 == 0 ve Part2 (Part1 ^ 0x1337) + 42

ludzof on 2026-07-28 21:41: this level has a great mathematical trap. The program uses two checks to validate your keys: Key Part 1 must be a multiple of 17. (The strange mul F0F0F0F1 instruction is actually compiler-optimized modulo division for 17). Key Part 2 must equal (Key Part 1 ^ 0x1337) + 0x2A. If you just find any valid pair (like 17 and 4944), you'll bypass the checks and get the "LICENSE VALIDATED!" message, but the flag will decrypt into complete garbage. This is because the XOR key used to decrypt the hidden array is strictly the lowest byte of Key Part 2. By taking the gibberish output and reverse-XORing it, you can figure out the encrypted array in memory. Brute-forcing that array reveals the correct XOR key must be 0xBD. So, we just need a multiple of 17 that produces a Key 2 ending in 0xBD. The Perfect Solution: Key Part 1: 1700 Key Part 2: 5565 This mathematically satisfies both license checks AND perfectly decrypts the flag to Victory!.

sh4man on 2026-07-31 12:44: [Click to reveal]

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Solution by MoriartyPuth on 2026-07-15 18:20:

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crackmes.one

crackmes.one

Lacks's Level 5